What is the general solution for 2-sin^2(x)=1+cos(x) ?

The general solution for 2-sin^2(x)=1+cos(x) is x=2pin,x= pi/2+2pin,x=(3pi)/2+2pin

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2-sin^2(x)=1+cos(x)

2 - sin^{2} (x) = 1 + cos (x)

x = 2 πn, x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

Solution steps

2 - sin^{2} (x) = 1 + cos (x)

Subtract

1 + cos (x)

from both sides

- sin^{2} (x) - cos (x) + 1 = 0

Rewrite using trig identities

- cos (x) + cos^{2} (x) = 0

Solve by substitution

cos (x) = 1, cos (x) = 0

cos (x) = 1 : x = 2 πn

cos (x) = 0 : x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

Combine all the solutions

x = 2 πn, x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

2 sin^{2} (θ) - cos (θ) = 0

csc (θ) = \frac{2}{3}

cos (\frac{x}{2}) (\frac{1}{2}) = 0

cos (x) = 0.83

sin (x + \frac{π}{4}) = \frac{1}{2}, 0 \leq x \leq 2 π

What is the general solution for 2-sin^2(x)=1+cos(x) ?
The general solution for 2-sin^2(x)=1+cos(x) is x=2pin,x= pi/2+2pin,x=(3pi)/2+2pin